If Δ 1 = x s i n θ c o s θ - s i n θ - x 1 c o s θ 1 x and Δ 2 = x s i n 2 θ c o s 2 θ - s i n 2 θ - x 1 c o s 2 θ 1 x , x ≠ 0 ; then for all θ ∈ 0 ,   π 2 :

Question

If &#916; 1 = x s i n &#952; c o s &#952; - s i n &#952; - x 1 c o s &#952; 1 x and &#916; 2 = x s i n 2 &#952; c o s 2 &#952; - s i n 2 &#952; - x 1 c o s 2 &#952; 1 x , x &#8800; 0 ; then for all &#952; &#8712; 0 , &#8201; &#960; 2 :

Aicharya · Accepted Answer

&#916; 1 = x s i n &#952; - s i n &#952; - x c o s &#952; 1 &#160; &#160; &#160; &#160; c o s &#952; 1 x = x - x 2 - 1 - s i n &#952; - x s i n &#952; - c o s &#952; + c o s &#952; ( - s i n &#952; + x c o s &#952; ) = - x 3 - x + x s i n 2 &#952; + s i n &#952; cos &#8289; &#952; - s i n &#952; c o s &#952; + x c o s 2 &#952; = - x 3 - x + x sin 2 &#8289; &#952; + cos 2 &#8289; &#952; = - x 3 Similarly: &#916; 2 = - x 3 &#8756; &#916; 1 + &#916; 2 = - 2 x 3

If Δ 1 = x s i n θ c o s θ - s i n θ - x 1 c o s θ 1 x and Δ 2 = x s i n 2…

Solution

If &#916; 1 = x s i n &#952; c o s &#952; - s i n &#952; - x 1 c o s &#952; 1 x and &#916; 2 = x s i n 2…

Solution

If Δ 1 = x s i n θ c o s θ - s i n θ - x 1 c o s θ 1 x and Δ 2 = x s i n 2…